Surface Integrals: Finding Centroid & Inertia of Circle

In summary, to find the centroid of a circle with surface-density varying as the nth power of the distance from a point O on the circumference, we can use polar coordinates and calculate the integral of x r^n dx dy over the disk, divided by the mass. This can be done by assuming O is the origin and the circle is (x-1)^2+y^2=1. The moment of inertia calculations follow a similar process.
  • #1
CaptainX
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How to find the centroid of circle whose surface-density varies as the nth power of the distance from a point O on the circumference. Also it's moments of inertia about the diameter through O.

I'm getting x'=2a(n-2)/(n+2)
And about diameter
-4(a×a)M{something}
 
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  • #2
Working in coordinates, we can assume that ##O## is the origin, and the circle is ##(x-1)^2+y^2=1##. In polar coordinates, this circle is ##r=2\cos(\theta)##. The ##y##-coordinate of the center of mass is ##0## by symmetry, and the ##x##-coordinate is the integral of ##x r^n dx dy=\cos\theta \ r^{n+2} dr d\theta ## over the disk, divided by the mass, which should be straightforward to calculate (bounds are ##-\pi/2\leq\theta\leq\pi/2## and ##0\leq r\leq 2\cos\theta##).

The moment of inertia calculations are similar.
 
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  • #3
Infrared said:
Working in coordinates, we can assume that ##O## is the origin, and the circle is ##(x-1)^2+y^2=1##. In polar coordinates, this circle is ##r=2\cos(\theta)##. The ##y##-coordinate of the center of mass is ##0## by symmetry, and the ##x##-coordinate is the integral of ##x r^n dx dy=\cos\theta \ r^{n+2} dr d\theta ## over the disk, divided by the mass, which should be straightforward to calculate (bounds are ##-\pi/2\leq\theta\leq\pi/2## and ##0\leq r\leq 2\cos\theta##).

The moment of inertia calculations are similar.
Thanks a lot!
 

What is a surface integral?

A surface integral is a mathematical operation that calculates the total value of a function over a two-dimensional surface. It is similar to a regular integral, but instead of integrating over a one-dimensional interval, it integrates over a two-dimensional surface.

How do you find the centroid of a circle using surface integrals?

The centroid of a circle can be found by using the formula:
Cx = (1/A) * ∫∫ x * f(x,y) dA
Cy = (1/A) * ∫∫ y * f(x,y) dA
where A is the total area of the circle and f(x,y) is the function representing the circle. This formula calculates the average x and y coordinates of the circle, which is the centroid.

What is inertia and how is it related to surface integrals?

Inertia is a measure of an object's resistance to changes in its motion. In the context of surface integrals, inertia is used to calculate the moment of inertia of a circle. The moment of inertia is a measure of an object's rotational inertia, and it can be calculated by using surface integrals.

Can surface integrals be used to find the moment of inertia of any shape?

Yes, surface integrals can be used to find the moment of inertia of any shape as long as the shape can be represented by a function in two-dimensional space. This includes not only circles, but also other shapes such as squares, triangles, and more complex shapes.

What are some real-world applications of surface integrals in finding centroids and inertia?

Surface integrals are commonly used in engineering and physics to calculate the center of mass and moment of inertia of objects. This information is important in designing structures, vehicles, and other systems that need to have a stable and balanced distribution of weight. Surface integrals are also used in fluid mechanics to calculate the flow rate and pressure of fluids over a surface.

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